Problem 91 - A Complete Resource Book in Chemistry for JEE Main

Question

During the decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) to give oxygen, \(48 \mathrm{~g} \mathrm{O}_{2}\) is formed per minute at a certain point of time. The rate of formation of water at this point is (a) \(0.75 \mathrm{~mol} \mathrm{~min}^{1}\) (b) \(1.5 \mathrm{~mol} \mathrm{~min}^{-1}\) (c) \(2.25 \mathrm{~mol} \mathrm{~min}^{-1}\) (d) \(3.0 \mathrm{~mol} \mathrm{~min}^{-1}\)

Step-by-Step Solution

Answer

The rate of formation of water is \(3.0 \mathrm{~mol} \mathrm{~min}^{-1}\).

Step 1: Write the Balanced Equation

The chemical reaction for the decomposition of hydrogen peroxide (\(\mathrm{H_2O_2}\)) is balanced as follows: \[2 \mathrm{H_2O_2} \rightarrow 2 \mathrm{H_2O} + \mathrm{O_2}\]. This equation tells us that for every 2 moles of \(\mathrm{H_2O_2}\) that decompose, 1 mole of \(\mathrm{O_2}\) and 2 moles of \(\mathrm{H_2O}\) are produced.

Step 2: Calculate Moles of \(\mathrm{O_2}\) Formed

Given that \(48 \text{ g of } \mathrm{O_2}\) is formed per minute, we need to first calculate the number of moles of \(\mathrm{O_2}\) formed. The molar mass of \(\mathrm{O_2}\) is \(32 \text{ g/mol}\). \[\text{Moles of } \mathrm{O_2} = \frac{48 \text{ g}}{32 \text{ g/mol}} = 1.5 \text{ moles/min}\].

Step 3: Determine Moles of \(\mathrm{H_2O}\) Formed

According to the balanced equation, for every 1 mole of \(\mathrm{O_2}\) formed, 2 moles of \(\mathrm{H_2O}\) are formed. Therefore, if \(1.5 \text{ moles/min of } \mathrm{O_2}\) are formed, then the rate of formation of \(\mathrm{H_2O}\) is:\[\text{Rate of formation of } \mathrm{H_2O} = 1.5 \times 2 = 3.0 \text{ moles/min}\].

Key Concepts

Balanced Chemical Equation

In any chemical reaction, a balanced chemical equation is a must-have tool for understanding the quantities of reactants and products involved. The decomposition of hydrogen peroxide (\(\mathrm{H_2O_2}\)) is expressed in a balanced equation as: \[2 \mathrm{H_2O_2} \rightarrow 2 \mathrm{H_2O} + \mathrm{O_2}\]This tells us the precise relationship: every 2 moles of \(\mathrm{H_2O_2}\) break down to give 2 moles of water (\(\mathrm{H_2O}\)) and 1 mole of oxygen gas (\(\mathrm{O_2}\)).
When we balance an equation, we ensure that the number of atoms for each element is equal on both sides of the reaction. This follows the Law of Conservation of Mass.

• Helps us predict the products created.
• Enables calculations of reactants needed and products formed.

This is crucial for determining how much of each substance participates or forms during the reaction.

Stoichiometry

Stoichiometry is the magic that lets chemists translate between the quantities of different substances involved in a reaction. It's like the blueprint that connects a balanced chemical equation to real-world lab calculations.In stoichiometry, we use the ratios from a balanced equation to figure out and predict different quantities, such as:

• How much product will form.
• How much reactant is needed.

For our example, the decomposition of hydrogen peroxide tells us that the formation of 1 mole of \(\mathrm{O_2}\) releases 2 moles of \(\mathrm{H_2O}\).
This gives us a clear stoichiometric ratio of: \(1\,\mathrm{mol}\,\mathrm{O}_2:2\,\mathrm{mol}\,\mathrm{H}_2\mathrm{O}\).
This understanding is vital to convert between units or substances during further calculations.

Mole Calculations

Moles connect the microscopic world of atoms to the macroscopic world we see in labs. Calculations involving moles allow us to measure and describe how much of a substance is involved in a reaction.When given the mass of a substance, such as \(\mathrm{O_2}\) formed, mole calculations allow us to convert it to moles:1. Calculate moles using the formula: \[ \text{Moles} = \frac{\text{Mass in grams}}{\text{Molar mass in g/mol}} \]2. For our example: Given \(48 \,\mathrm{g}\) of \(\mathrm{O_2}\), which has a molar mass of \(32 \,\mathrm{g/mol}\), the moles of \(\mathrm{O_2}\) is: \[ \frac{48 \text{ g}}{32 \text{ g/mol}} = 1.5 \text{ moles} \]
Understanding moles is like having the key to the atomic-scale world, as it facilitates precise conversions and reactions as we predict and measure quantities involved in chemical processes.

Reaction Rate

The reaction rate tells us how fast or slow a reaction occurs. It's like the speedometer of the chemical process. It can be expressed as how much product is formed or reactant consumed over time.
In our case, the formation rate of \(48\, \mathrm{g/min}\) of \(\mathrm{O_2}\). Here, we can calculate the rate in terms of moles of product formed per minute using our previous calcualtions.For the decomposition of \(\mathrm{H_2O_2}\):

• \(\mathrm{O}_2\) forms at a rate of \(1.5\, \mathrm{moles/min}\).
• The balanced equation tells us each mole of \(\mathrm{O_2}\) corresponds with \(2\) moles of \(\mathrm{H_2O}\) forming.

So, \(\mathrm{H_2O}\) forms at \(3.0\, \mathrm{moles/min}\). This conversion allows us to explore reaction rates properly and gives insights into how effectively a reaction proceeds in real-time.